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If the medians of a triangle ABC interse...

If the medians of a `triangle ABC` intersect at `G`, show that
`ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)`
`=(1)/(3)ar(triangleABC)`.

Text Solution

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GIVEN:
A `triangleABC`. Its medians `AD, BE` and `CF` intersect at `G`.
TO PROVE `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC)`
PROOF:
We know that a median of a triangle divides it into two triangle of equal area.
In `triangleABC` ,`AD` is the median.
`therefore ar(triangleABD)=ar(triangleACD)." "...(i)`
In `triangleGBC`, `GD` is the median.
`therefore ar(triangleGBD)=ar(triangleGCD)." "...(ii)`
From (i) and (ii), we get
`ar(triangleABD)-ar(triangleGBD)=ar(triangleACD)-ar(triangleGCD)`
`rArr ar(triangleAGB)=ar(triangleAGC)`.
Similarly, `ar(triangleAGB)=ar(triangleBGC)`.
`rArr ar(triangleAGB)=ar(triangleAGC)`.
But, `ar(triangleABC)=ar(triangleAGB)+ar(triangleAGC)+ar(triangleBGC)`
`=3ar(triangleAGB)" "["using (iii)"]`.
`therefore ar(triangleAGB)=(1)/(3)ar(triangleABC)`.
Hence, `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC).`
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